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  • Explain Solution R.D.Sharma Class 12 Chapter 30 Probability Exercise 30.4 Question 21 Maths Textbook Solution.

Explain Solution R.D.Sharma Class 12 Chapter 30 Probability  Exercise 30.4 Question 21 Maths Textbook Solution.

Answers (1)

Answer:  i. \frac{16}{81}   ii. \frac{20}{81}   iii. \frac{40}{81} 

Hint: Probability=\frac{No.\: of\: outcomes}{Total\: outcomes}

Given: Two balls are drawn at random with replacement from a box containing 10 black and 8 red balls.

Solution:

\begin{aligned} &\text { Total balls }=10 \text { black }+8 \text { red }=18 b a l l s \\ \end{aligned}

\begin{aligned} &P(\text { first red ball })=\frac{8}{18} \\ \end{aligned}

\begin{aligned} &\left.P(\text { second red ball })=\frac{8}{18} \ldots \text { fas replacement is allowed }\right\} \\ \end{aligned}

\begin{aligned} &P(\text { first ball is black })=\frac{10}{18} \\ &P(\text { sec ond ball is black })=\frac{10}{18} \ldots\{\text { as replacement is allowed }\} \end{aligned}

i. \quad P( two red balls )=\frac{8}{18} \times \frac{8}{18}=\frac{16}{81}

ii. \quad \mathrm{P}( first \: ball\: is \: black \: and\: second \: is \: red )=\frac{10}{18} \times \frac{8}{18}=\frac{20}{81}

iii. \mathrm{P} (one of them is black and other is red) =\mathrm{P} (first ball is red \& second is black)

                                                                               +\mathrm{P}( first ball is black \& second is red)

                                                                            \begin{aligned} &=\frac{8}{18} \times \frac{10}{18}+\frac{10}{18} \times \frac{8}{18} \\ &=\frac{40}{81} \end{aligned}

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